[LeetCode]题解（python）：126-Word Ladder II

题目来源：

　　https://leetcode.com/problems/word-ladder-ii/

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题意分析：

　　给定一个beginWord和一个endWord，以及一个字典单词，找出所有从beginWord变成endWord的所有最短路径，单词变化每次只能变一个字母，所变的字母必须在字典里面。

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题目思路：

　　这是一个最短路径问题。首先将beginWord放进队列，当队列不为空的时候，pop出第一个数，将它周围的在字典的push进队列。要注意的是将字段的数push进队列的同时，将其路径用dict记录下来。

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代码（python）：

　　

class Solution(object):
    def findLadders(self, beginWord, endWord, wordlist):
        """
        :type beginWord: str
        :type endWord: str
        :type wordlist: Set[str]
        :rtype: List[List[int]]
        """
        ans,q = {},[]
        q.append(beginWord)
        ans[beginWord] = [[beginWord]]
        ans[endWord] = []
        while len(q) != 0:
            tmp = q.pop(0)
            for i in range(len(beginWord)):
                part1,part2 = tmp[:i],tmp[i + 1:]
                for j in "abcdefghijklmnopqrstuvwxyz":
                    if tmp[i] != j:
                        newword = part1 + j + part2
                        if newword == endWord:
                            for k in ans[tmp]:
                                ans[endWord].append(k + [endWord])
                            while len(q) != 0:
                                tmp1 = q.pop(0)
                                if len(ans[tmp1][0]) >= len(ans[endWord][0]):
                                    break
                                for ni in range(len(beginWord)):
                                    npart1,npart2 = tmp1[:ni],tmp1[ni+1:]
                                    for nj in "abcdefghijklmnopqrstuvwxyz":
                                        if tmp1[ni] != nj:
                                            nw = npart1 + nj + npart2
                                            if endWord == nw:
                                                for nk in ans[tmp1]:
                                                    ans[endWord].append(nk + [endWord])
                            break
                        if newword in wordlist:
                            q.append(newword)
                            ans[newword] = []
                            for k in ans[tmp]:
                                ans[newword].append(k + [newword])
                            wordlist.remove(newword)
                        elif newword in ans and len(ans[newword][0]) == len(ans[tmp][0]) + 1:
                            for k in ans[tmp]:
                                ans[newword].append(k + [newword])
        return ans[endWord]